A thin metal disc of radius 'r' floats on the water surface and bends the surface downwards along the perimeter, making an angle '$\theta$' with the vertical edge of the disc. If the weight of water displaced by the disc is 'W', the weight of the metal disc is
[ T = surface tension of water]

For the thin metal disc floating on the water surface, the forces acting on it in the vertical direction are in equilibrium.

The downward force is the weight of the metal disc, $W_{\text{disc}}$.

The upward forces are:
1. The buoyant force, which is equal to the weight of the water displaced by the disc, given as $W$.
2. The force due to surface tension. Since the water surface is bent downwards, the surface tension pulls the disc upwards. The surface tension force acts tangentially to the liquid surface, making an angle $\theta$ with the vertical edge of the disc.

The upward component of the surface tension force per unit length is $T \cos\theta$.
The total upward force due to surface tension along the perimeter of the disc is $2\pi r T \cos\theta$.

Equating the downward and upward forces for equilibrium:
$W_{\text{disc}} = W + 2\pi r T \cos\theta$

Answer: $2\pi r\text{T}\cos\theta + \text{W}$